Question

Let X denote the reaction time, in seconds, to a certain stimulus and Y denote the temperature (◦F) at which a certain reaction starts to take place. Suppose that two random variables X and Y have the joint density f(x,y)= 4xy, 0

Answer 1

Answer:

the probability that 0≤x≤1/2 and 1/4≤y≤1/2 is 3/64 (4.68%)

Step-by-step explanation:

assuming that X and Y are independent variables for the probability density function f(x,y) :

f(x,y) = 4*x*y  for 0≤x≤1 and 0≤y≤0

f(x,y) = 0 elsewhere

then the probability is calculated through:

P(x,y)= ∫f(x,y) dx dy = ∫4xy dx dy

for 0≤x≤1/2 and   1/4≤y≤1/2 we have

P(0≤x≤1/2,1/4≤y≤1/2 ) = ∫4xy dx dy = ∫4xy dx dy = 4*∫x dx ∫y dx = 4*[((1/2)²/2-0²/2)] *[(1/2)²/2-(1/4)²/2)] = 1 * 1/4 * (1/4-1/16) = 1 * 1/4 * 3/16 = 3/64

then the probability that 0≤x≤1/2 and 1/4≤y≤1/2 is 3/64 (4.68%)

Answer 2

The probability that 0≤x≤1/2 and 1/4≤y≤1/2 is 0.046875

How to determine the probability?

The distribution function is given as:

f(x,y) = 4xy 0≤x≤1 and 0≤y≤0

The probability is calculated using the following integral

P(x,y)= ∫f(x,y) dx dy = ∫4xy dx dy

So, we have:

$P(x,y) = \int\limits^{1/2}_{1/4} {\int\limits^{1/2}_{0} 4xy} \ dx dy$

Factor out 4

$P(x,y) = 4\int\limits^{1/2}_{1/4} {\int\limits^{1/2}_{0} xy} \ dx dy$

Integrate with respect to x

$P(x,y) = 4\int\limits^{1/2}_{1/4} { \frac{x^2}{2}y|\limits^{1/2}_{0}} dy$

Expand

$P(x,y) = 4\int\limits^{1/2}_{1/4} { \frac{1/2^2 - 0^2}{2}y dy$

$P(x,y) = 4\int\limits^{1/2}_{1/4} { \frac{1}{8}y\ dy$

Factor out 1/8

$P(x,y) = \frac{1}{2}\int\limits^{1/2}_{1/4} { y\ dy$

Factorize

$P(x,y) = \frac{1}{2} * \frac{y^2}{2}|\limits^{1/2}_{1/4}$

Expand

$P(x,y) = \frac{1}{2} * \frac{1/2^2 - 1/4^2}{2}$

$P(x,y) = \frac{0.1875}{4}$

Evaluate the quotient

P(x,y) = 0.046875

Hence, the probability that 0≤x≤1/2 and 1/4≤y≤1/2 is 0.046875

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